Solveeit Logo
Login

Question

Physics Question on simple harmonic motion

The piston in the cylinder head of a locomotive has a stroke of 6 m6 \,m. If the piston executing simple harmonic motion with an angular frequency of 200 rad min−1200 \,rad\, min^{-1} , its maximum speed is

A

5 m s−15\, m\,s^{-1}

B

10 m s−110\, m\,s^{-1}

C

15 m s−115\, m\,s^{-1}

D

20 m s−120\, m\,s^{-1}

Answer

10 m s−110\, m\,s^{-1}

Explanation

Solution

Here, ω=200 rad min−1\omega = 200 \,rad\, min^{-1} Stroke means twice the amplitude. ∴A=3 m\therefore A = 3\, m, Maximum speed, vmax=Aωv_{max} = A\omega =3×200=600 m rad min−1= 3 \times200 = 600 \,m \,rad \,min ^{-1} =60060 m s−1= \frac{600}{60} \,m \,s^{-1} =10 m s−1 = 10 \,m \,s^{-1}